Group Homomorphisms Review

Group Homomorphisms Review

We will now review some of the recent material regarding group homomorphisms.

  • On the Group Homomorphisms page we said that two groups $(G_1, *_1)$ and $(G_2, *_2)$ are Homomorphic if there exists a function $f : G_1 \to G_2$ called a Homomorphism which has the following properties:
  • * (1) For all $x, y \in G_1$ we have that:
(1)
\begin{align} \quad f(x *_1 y) = f(x) *_2 f(y) \end{align}
  • Note that the only difference between a group isomorphism and a group homomorphism is that group isomorphisms require that $f$ be bijective. Group homomorphisms do not!
Theorem
(a) If $(G_1, *_1)$ and $(G_2, *_2)$ are homomorphic with homomorphism $f : G_1 \to G_2$ and if $e$ is the identity in $G_1$ then $f(e)$ is the identity in $G_2$.
(b) If $(G_1, *_1)$ and $(G_2, *_2)$ are homomorphic with homomorphism $f : G_1 \to G_2$ and if $x \in G_1$ has inverse $x^{-1} \in G_1$ then $f(x) \in G_2$ has inverse $f(x^{-1}) \in G_2$.
(c) If $(G_1, *_1)$ and $(G_2, *_2)$ are homomorphic with homomorphism $f : G_1 \to G_2$ and if $H \subseteq G_1$ is a subgroup of $G_1$ then $f(H) \subseteq G_2$ is a subgroup of $G_2$.
(d) If $(G_1, *_1)$ and $(G_2, *_2)$ are homomorphic with homomorphism $f : G_1 \to G_2$ and if $H \subseteq G_2$ is a subgroup of $G_2$ and $f^{-1}(H) \neq \emptyset$ then $f^{-1}(H)$ is a subgroup of $G_1$.
  • On The Kernel of a Group Homomorphism page we said that if $(G_1, *_1)$ and $(G_2, *_2)$ are homomorphic with homomorphism $f : G_1 \to G_2$ where $e_2$ is the identity of $G_2$ then the Kernel of $f$ is defined as:
(2)
\begin{align} \quad \mathrm{ker} (f) = \{ x \in G_1 : f(x) = e_2 \} \end{align}
  • That is, $\mathrm{ker} (f)$ is the set of all elements in $G_1$ that are mapped to the identity element in $G_2$.
  • We noted that $\mathrm{ker} (f)$ is a subgroup of $G_1$.
  • We then proved an important result:
Theorem
(a) If $(G_1, *_1)$ and $(G_2, *_2)$ are homomorphic with homomorphism $f : G_1 \to G_2$ and with identities $e_1$ and $e_2$ respectively, if $\mathrm{ker} (f) = \{ e_1 \}$ then $f$ is injective.
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